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X->0 (sin[x])/x

by phymatter
Tags: sinx or x, x>0
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phymatter
#1
Mar19-11, 05:20 AM
P: 131
does lim x->0 (sin[x])/x exist ? if yes then what is it , iguess 0 , but cannot figure out the reason .. pl. help...
note: [x] is greatest integer less than or equal to x
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tiny-tim
#2
Mar19-11, 06:21 AM
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hi phymatter!
Quote Quote by phymatter View Post
does lim x->0 (sin[x])/x exist ? if yes then what is it , iguess 0
yes

(just choose your delta to be 0.9, whatever your epsilon )
mathman
#3
Mar19-11, 03:37 PM
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If-1 < x < 0, [x] = -1, if 0 < x < 1, [x] = 0. As a result , for x < 0, the limit is -∞ while for x > 0, the limit is 0.

TylerH
#4
Mar19-11, 05:40 PM
P: 737
X->0 (sin[x])/x

Quote Quote by phymatter View Post
note: [x] is greatest integer less than or equal to x
It's usually called the "floor" function.

I don't think it would exist. lim(x->0+) sin(floor(x))/x = 0 and lim(x->0-) sin(floor(x))/x = ∞. For there to be a limit, lim(x->0+) sin(floor(x))/x and lim(x->0-) sin(floor(x))/x must be equal, they're not.


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